(26x-4)(5x)=120

Solution for (26x-4)(5x)=120 equation:

(26x-4)(5x)=120

We move all terms to the left:

(26x-4)(5x)-(120)=0

We multiply parentheses

130x^2-20x-120=0

a = 130; b = -20; c = -120;
Δ = b2-4ac
Δ = -202-4·130·(-120)
Δ = 62800
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{62800}=\sqrt{400*157}=\sqrt{400}*\sqrt{157}=20\sqrt{157}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-20\sqrt{157}}{2*130}=\frac{20-20\sqrt{157}}{260}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+20\sqrt{157}}{2*130}=\frac{20+20\sqrt{157}}{260}
$